# Introduction

We derive in this document a first version of Cauchy’s integral theorem:

### Theorem – Cauchy’s Integral Theorem (Local Version).

‌ Let $$f: \Omega \to \mathbb{C}$$ be a holomorphic function. For any $$a \in \Omega,$$ there is a radius $$r>0$$ such that the open disk $$D(a, r)$$ is included in $$\Omega$$ and for any rectifiable closed path $$\gamma$$ of $$D(a, r),$$ $\int_{\gamma} f(z) \, dz = 0.$
We will actually state and prove a slightly stronger version – one that does not require the restriction to small disks if $$\Omega$$ is star-shaped.

In a subsequent document, we will prove an even more general result, the global version of Cauchy’s integral theorem. It will be applicable if $$\Omega$$ is merely simply connected (that is “without holes”).

# Integral Lemma for Polylines

### Lemma – Integral Lemma for Triangles.

‌ Let $$f:\Omega \to \mathbb{C}$$ be a holomorphic function. If $$\Delta$$ is a triangle with vertices $$a,$$ $$b$$ and $$c$$ which is included in $$\Omega$$ $\Delta = \{ \lambda a + \mu b + \nu c \; | \; \lambda \geq 0, \, \mu \geq 0, \, \nu \geq 0 \, \mbox{ and } \, \lambda + \mu + \nu = 1 \} \subset \Omega$ and if $$\gamma = [a \to b \to c \to a]$$ is an oriented boundary of $$\Delta$$ then $\int_{\gamma} f(z) \, dz = 0.$

### Proof.

‌ Let $$a_0=a,$$ $$b_0=b,$$ $$c_0=c;$$ consider the midpoints of the triangle edges: $d_0 = \frac{b_0 + c_0}{2}, \; e_0 = \frac{a_0 + c_0}{2}, \; f_0 = \frac{a_0 + b_0}{2}.$ The sum of the integrals of $$f$$ along the four paths $$[a_0 \to f_0 \to e_0 \to a_0],$$ $$[f_0 \to b_0 \to d_0 \to f_0],$$ $$[e_0 \to d_0 \to c_0 \to e_0],$$ $$[d_0 \to e_0 \to f_0 \to d_0]$$ is equal to the integral of $$f$$ along $$\gamma.$$ By the triangular inequality, there is at least one path in this set, that we denote $$\gamma_1,$$ such that $\left| \int_{\gamma_1} f(z) \, dz \right| \geq \frac{1}{4} \left| \int_{\gamma} f(z) \, dz \right|.$ We can iterate this process and come up with a sequence of paths $$\gamma_n$$ such that $\left| \int_{\gamma_n} f(z) \, dz \right| \geq \frac{1}{4^n} \left| \int_{\gamma} f(z) \, dz \right|.$ Denote $$\Delta_n$$ the triangles associated to the $$\gamma_n;$$ they form a sequence of non-empty and nested compact sets. By Cantor’s intersection theorem, there is a point $$w$$ such that $$w \in \Delta_n$$ for every natural number $$n.$$ The differentiability of $$f$$ at $$w$$ provides a complex-valued function $$\epsilon_w,$$ defined in a neighbourhood of $$0,$$ such that $$\lim_{h \to 0} \epsilon_w(h) = \epsilon_w(0) = 0$$ and $f(z) = f(w) + f'(w) (z - w) + \epsilon_w(z - w) |z-w|$ Consequently, for any $$\epsilon > 0$$ and for any number $$n$$ large enough, $\left| \int_{\gamma_n} [f(z) - f(w) - f'(w) (z - w)] \, dz\right| \leq \epsilon \, \mathrm{diam} \, \Delta_n \times \ell(\gamma_n),$ where the diameter of a subset $$A$$ of the complex plane is defined as $\mathrm{diam} \, A = \sup \, \{|z - w| \; | \; z \in A, \, w \in A\}.$ We have $$\ell(\gamma_n) = \ell(\gamma)/2^n$$ and $$\mathrm{diam} \, \Delta_n = \mathrm{diam} \, \Delta_0 / 2^n.$$ Additionally, $\int_{\gamma_n} f(w) \,dz = \int_{\gamma_n} f'(w) (z - w) \, dz = 0$ since the functions $$z \in \mathbb{C} \mapsto f(w)$$ and $$z \in \mathbb{C} \mapsto f'(w) (z-w)$$ have primitives. Consequently, for any $$\epsilon > 0,$$ for $$n$$ large enough, $\frac{1}{4^n} \left| \int_{\gamma} f(z) \, dz \right| \leq \left| \int_{\gamma_n} f(z) \, dz\right| \leq \frac{1}{4^n}\epsilon \, \mathrm{diam} \, \Delta_0 \times \ell(\gamma),$ which is only possible if the integral of $$f$$ along $$\gamma$$ is zero. $$\blacksquare$$

### Definition – Star-Shaped Set.

‌ A subset $$A$$ of the complex plane is star-shaped if it contains at least one point $$c$$ – a (star-)center, the set of which is called the kernel of $$A$$ – such that for any $$z$$ in $$A,$$ the segment $$[c, z]$$ is included in $$A.$$

### Lemma – Integral Lemma for Polylines.

‌ Let $$f: \Omega \to \mathbb{C}$$ be a holomorphic function where $$\Omega$$ is an open star-shaped subset of $$\mathbb{C}.$$ For any closed path $$\gamma = [a_0 \to \dots \to a_{n-1} \to a_0]$$ of $$\Omega,$$ $\int_{\gamma} f(z) \, dz = 0.$

### Proof.

‌ Let $$c$$ be a star-center of $$\Omega$$ and define $$a_n = a_0;$$ for any $$k \in \{0, \dots, n-1\},$$ the triangle with vertices $$c,$$ $$a_{k}$$ and $$a_{k+1}$$ is included in $$\Omega.$$ Hence, by the integral lemma for triangles, the integral along the path $$\gamma_{k} = [c \to a_{k} \to a_{k+1} \to c]$$ of $$f$$ is zero. Now, as $\int_{\gamma} f(z) \, dz = \sum_{k=0}^{n-1} \int_{\gamma_{k}} f(z) \, dz,$ the integral of $$f$$ along $$\gamma$$ is zero as well. $$\blacksquare$$

# Approximations of Rectifiable Paths by Polylines

To extend the integral lemma beyond closed polylines, we prove that polylines provide appropriate approximations of rectifiable paths:

### Lemma – Polyline Approximations of Rectifiable Paths.

‌ Let $$\gamma$$ be a rectifiable path. For any $$\epsilon_{\ell} > 0$$ and $$\epsilon_{\infty} > 0,$$ there is an oriented polyline $$\mu,$$ with the same endpoints as $$\gamma,$$ such that $\ell(\mu - \gamma) \leq \epsilon_{\ell} \; \mbox{ and } \; \forall \; t \in [0,1], \, |(\mu - \gamma)(t)| \leq \epsilon_{\infty}.$

### Proof – Polyline Approximations of Rectifiable Paths.

‌ Suppose that the path $$\gamma$$ is continuously differentiable. Let $$(t_0, \dots, t_n)$$ be a partition of the interval $$[0,1]$$ and let $$\mu$$ be the associated polyline: $\mu = [\gamma(t_0) \to \gamma(t_1)] \, |_{t_1} \, \cdots \, |_{t_{n-1}} \, [\gamma(t_{n-1}) \to \gamma(t_n)]$ The path $$\gamma$$ and $$\mu$$ have the same endpoints. The path $$\gamma$$ may be considered as the concatenation $$\gamma = \gamma_1 \, |_{t_1} \, \dots \, |_{t_{n-1}} \, \gamma_n$$ with the paths $$\gamma_{k}$$ defined by $\forall \, k \in \{1, \dots, n\}, \, \forall t \in [0,1], \, \gamma_k(t) = \gamma\left(t_{k-1} + t (t_{k} - t_{k-1})\right),$ hence we have $\ell(\mu - \gamma) = \sum_{k=1}^{n} \int_0^1 \left| \gamma(t_{k}) - \gamma(t_{k-1}) - \gamma'_k(t)\right| \, dt.$ As $\gamma(t_{k}) - \gamma(t_{k-1}) = \int_{t_{k-1}}^{t_{k}} \gamma'(s) \, ds$ and $\begin{split} \gamma'_k(t) &= (t_{k} - t_{k-1})\gamma'\left(t_{k-1} + t (t_{k} - t_{k-1})\right) \\ &= \int_{t_{k-1}}^{t_{k}} \gamma'\left(t_{k-1} + t (t_{k} - t_{k-1})\right) \, ds, \end{split}$ we have the inequality $\ell(\mu - \gamma) \leq \int_0^1 \left[\sum_{k=1}^{n} \int_{t_{k-1}}^{t_{k}} |\gamma'(s) - \gamma'(t_{k-1} + t (t_{k} - t_{k-1}))| \,ds \right] dt$ The function $$\gamma'$$ is by assumption continuous, and hence uniformly continuous, on $$[0,1],$$ therefore for any $$\epsilon > 0,$$ there is a $$\delta(\epsilon) > 0$$ such that, $$|\gamma'(s) - \gamma'(t)| < \epsilon$$ whenever $$|s - t| < \delta(\epsilon).$$ For any $$\epsilon_{\ell} > 0,$$ for any partition $$(t_0,\dots, t_n)$$ such that $$|t_{k} - t_{k-1}| < \delta(\epsilon_{\ell})$$ for any $$k \in \{1, \dots, n\},$$ we have $\ell(\mu - \gamma) \leq \int_0^1 \left[\sum_{k=1}^{n} \int_{t_{k-1}}^{t_{k}} \epsilon_{\ell} \,ds \right] dt = \epsilon_{\ell}.$ For any $$\epsilon_{\infty} > 0,$$ as $\forall \, t \in [0,1],\, |\mu(t) - \gamma(t)| \leq |\mu(0) - \gamma(0)| + \ell(\mu - \gamma) = \ell(\mu - \gamma),$ any partition $$(t_0,\dots, t_n)$$ such that $$|t_{k} - t_{k-1}| < \delta(\epsilon_{\infty})$$ ensures that $\forall \; t \in [0,1], \, |(\mu - \gamma)(t)| \leq \epsilon_{\infty}.$ If $$\gamma$$ is merely rectifiable, the same approximation process, applied to each of its continuously differentiable components provides the result. $$\blacksquare$$

# Cauchy’s Integral Theorem

We finally get rid of the polyline assumption:

### Theorem – Cauchy’s Integral Theorem (Star-Shaped Version).

‌ Let $$f: \Omega \to \mathbb{C}$$ be a holomorphic function where $$\Omega$$ is an open star-shaped subset of $$\mathbb{C}.$$ For any rectifiable closed path $$\gamma$$ of $$\Omega,$$ $\int_{\gamma} f(z) \, dz = 0.$

### Proof.

‌ Let $$\epsilon > 0.$$ Let $$r>0$$ be smaller than the distance between $$\gamma([0,1])$$ and $$\mathbb{C} \setminus \Omega.$$ The set $K = \{z \in \mathbb{C} \, | \, d(z, \gamma([0,1])) \leq r \},$ is compact and included in $$\Omega.$$ Consequently, the restriction of $$f$$ to $$K$$ is bounded and uniformly continuous: there is a $$M>0$$ such that $\forall \, z \in K, \; |f(z)| \leq M,$ and a there is a $$\eta_{\epsilon}>0$$ – smaller than or equal to $$r$$ – such that $\forall \, z \in K, \, \forall \, w \in \gamma([0,1]), \, |z - w| \leq \eta_{\epsilon} \, \Rightarrow \, |f(z) - f(w)| \leq \frac{\epsilon}{2 (\ell(\gamma)+1)}.$ Now, let $$\gamma_{\epsilon}$$ be a closed polyline approximation of $$\gamma$$ such that $\ell(\gamma_{\epsilon} - \gamma) \leq \frac{\epsilon}{2M} \; \mbox{ and } \; \forall \; t \in [0,1], \, |(\gamma_{\epsilon} - \gamma)(t)| \leq \eta_{\epsilon}.$ By construction, $$\gamma_{\epsilon}$$ belongs to $$K,$$ hence it is a closed path of $$\Omega.$$ Therefore, the integral lemma for polylines provides $\int_{\gamma_{\epsilon}} f(z) \, dz = 0.$

The rectifiable $$\gamma$$ and $$\gamma_{\epsilon}$$ have a decomposition into continuously differentiable paths associated to a common partition $$(t_0,\dots,t_n)$$ of the interval $$[0,1]$$: $\gamma = \gamma_1 \, |_{t_1} \, \cdots |_{t_n} \, \, \gamma_n \; \mbox{ and } \; \gamma_{\epsilon} = \gamma_{1\epsilon} \, |_{t_1} \, \cdots \, |_{t_n} \, \gamma_{n\epsilon}$

The difference between the integral of $$f$$ along $$\gamma$$ and $$\gamma_{\epsilon}$$ satisfies $\left| \int_{\gamma} f(z) \, dz - \int_{\gamma_{\epsilon}} f(z) \, dz \right| = \left| \sum_{k=1}^n \int_{0}^1 [ (f \circ \gamma_k) \gamma_k' - (f \circ \gamma_{\epsilon k}) \gamma_{\epsilon k}'](t) \, dt\right|$ Since for any $$k \in \{1,\dots, n\}$$ $(f \circ \gamma_k) \gamma_k' - (f \circ \gamma_{\epsilon k}) \gamma_{\epsilon k}' = (f \circ \gamma_k - f \circ \gamma_{\epsilon k}) \gamma'_{k} + (f \circ \gamma_{\epsilon k}) (\gamma_k' - \gamma_{\epsilon k}'),$ we have $\begin{split} &\left| \int_{\gamma} f(z) \, dz - \int_{\gamma_{\epsilon}} f(z) \, dz \right| \\ &\leq \, \left| \sum_{k=1}^n \int_{0}^1 [ (f \circ \gamma_k - f \circ \gamma_{\epsilon k}) \gamma'_{k} ](t) \, dt\right| \\ &+ \, \left| \sum_{k=1}^n \int_{0}^1 [ (f \circ \gamma_{\epsilon k}) (\gamma_k' - \gamma_{\epsilon k}') ](t) \, dt\right| \end{split}$ and thus $\begin{split} \left| \int_{\gamma} f(z) \, dz\right| \leq &\, \max_{t \in [0,1]} |f(\gamma(t)) - f(\gamma_{\epsilon}(t))| \times \ell(\gamma) \\ & \, + \max_{t \in [0,1]} |f(\gamma_{\epsilon}(t))| \times \ell(\gamma_{\epsilon} - \gamma) \\ \leq &\, \frac{\epsilon}{2 (\ell(\gamma)+1)} \times \ell(\gamma) + M \times \frac{\epsilon}{2 M} \\ \leq &\, \epsilon. \end{split}$ As $$\epsilon > 0$$ is arbitrary, the integral of $$f$$ along $$\gamma$$ is zero. $$\blacksquare$$

# Consequences

### Theorem – Cauchy’s Integral Formula for Disks.

‌ Let $$\Omega$$ be an open subset of the complex plane and $$\gamma = c + r[\circlearrowleft]$$ be an oriented circle such that the closed disk $$\overline{D}(c, r)$$ is included in $$\Omega.$$ For any holomorphic function $$f:\Omega \to \mathbb{C},$$ $\forall \, z \in D(c,r), \; f(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw.$

### Proof.

‌ Refer to the answers of exercise “Cauchy’s Integral Formula for Disks” $$\blacksquare$$

### Corollary – Derivatives are Complex-Differentiable.

‌ The derivative of any holomorphic function is holomorphic.

### Proof.

‌ Refer to the answers of exercise “Cauchy’s Integral Formula for Disks” $$\blacksquare$$

### Theorem – Morera’s Theorem.

‌ Let $$\Omega$$ be an open subset of $$\mathbb{C}.$$ A function $$f:\Omega \to \mathbb{C}$$ is holomorphic if and only if it is continuous and locally, its line integrals along rectifiable closed paths are zero: for any $$c \in \Omega,$$ there is a $$r>0$$ such that $$D(c,r) \subset \Omega$$ and for any rectifiable closed path $$\gamma$$ of $$D(c,r),$$ $\int_{\gamma} f(z) \, dz = 0.$

### Proof.

‌ If $$f$$ is holomorphic, then it is continuous and by Cauchy’s integral theorem, its line integrals along rectifiable closed paths are locally zero. Conversely, if $$f$$ is continuous and all its line integrals along closed rectifiable paths are zero in some non-empty open disk $$D(c,r)$$ of $$\Omega,$$ $$f$$ satisfies the condition for the existence of primitives in $$D(c,r).$$ Any such primitive is holomorphic; since derivatives are complex-differentiable its derivative is holomorphic too and $$f$$ is holomorphic in some neighbourhood of $$c.$$ Since the initial assumption holds for any $$c \in \Omega,$$ we can conclude that $$f$$ is holomorphic on $$\Omega.$$ $$\blacksquare$$

### Theorem – Limit of Holomorphic Functions.

‌ Let $$\Omega$$ be an open subset of $$\mathbb{C}.$$ If a sequence of holomorphic functions $$f_n: \Omega \to \mathbb{C}$$ converges locally uniformly to a function $$f:\Omega \to \mathbb{C},$$ that is if for any $$c \in \Omega,$$ there is a $$r>0$$ such that $$D(c,r) \subset \Omega$$ and $\lim_{n \to +\infty} \sup_{z \in D(c,r)} |f_n(z) - f(z)| = 0,$ then $$f$$ is holomorphic.

### Proof.

‌ The function $$f$$ is continuous as a locally uniform limit of continuous functions. Now, let $$c \in \Omega$$ and let $$r>0$$ be such that $$D(c,r) \subset \Omega$$ and the functions $$f_n$$ converge uniformly to $$f$$ in $$D(c,r).$$ By Cauchy’s integral theorem, for any rectifiable closed path $$\gamma$$ of $$D(c,r),$$ the integral of $$f_n$$ along $$\gamma$$ is zero. Thus $\int_{\gamma} f(z) \, dz = \lim_{ n\to +\infty} \int_{\gamma} f_n(z) \, dz = 0.$ By Morera’s theorem, $$f$$ is holomorphic. $$\blacksquare$$

### Theorem – Liouville’s Theorem.

‌ Any holomorphic function defined on $$\mathbb{C}$$ (any entire function) which is bounded is constant.

### Proof.

‌ Let $$f:\mathbb{C} \to \mathbb{C}$$ be a holomorphic function such that $$|f(z)| \leq \kappa$$ for any $$z \in \mathbb{C}.$$ Since derivatives are complex-differentiable, we may apply Cauchy’s integral formula for disks to the function $$f'$$ and to the oriented circle $$\gamma = z + r[\circlearrowleft]$$ for $$r>0$$ and $$z \in \mathbb{C}.$$ We have $f'(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f'(w)}{w-z} dw$ and by integration by parts, $f'(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{(w-z)^2} dw,$ which yields by the M-L inequality $|f'(z)| \leq \frac{\kappa}{r}.$ This inequality holds for any $$r>0,$$ thus $$f'(z) = 0$$. Consequently, the zero function and $$f$$ are both primitives of $$f'$$; since the domain of $$f'$$ is connected, these two primitives differ by a constant and thus $$f$$ is constant. $$\blacksquare$$